| Topic: |
Science > Physics |
| User: |
"Dave Typinski" |
| Date: |
08 Mar 2005 11:00:13 PM |
| Object: |
Decay chain physics? |
I'm trying to understand the energy released by (actually, the thermal
power of) nuclear decay chains so I can model geologic radioisotope
decay heat. Anyone care to point me in the right direction with
respect to nuclear decay energy release?
Yes, yes, I could look up all the relevant geothermodynamic data and
simply grind out an answer; heck, it's probably all been done already
anyway. However, I'm doing this as a recreational diversion and I
want to gain something along the way. Namely, a better understanding
of nuclear decay processes.
I've re-derived the simple equations for a three-step decay chain
that provide the daughter isotopes' quantities with respect to time.
For four step chains and further, I'll appeal to the Bateman
equation... and I haven't yet figured out how to derive that one. I
suppose it wouldn't hurt if I actually bothered to read Bateman's
paper...
Anyway, figuring the power output from any generation with respect to
time is easy *if* one knows the energy released per decay event. Yes,
I know about branching; that's easy to take into account.
Understanding the energy release attendant to a given decay event is
where I'm stuck. Obviously, I need a deeper understanding of nuclear
decay processes. I've searched a good deal online, searched the
_Modern Physics_ text used at UF, and I'm having trouble finding a
good in depth discussion of it all.
Take thorium-234, for example. It beta decays to protactinium-234.
The online table of the nuclides (at KAERI) lists it's decay energy as
273 keV. But, when I click on the "beta" in the page for Th-234, I
can't make heads or tails of the b- and gamma energy distribution
table and energy diagram.
http://atom.kaeri.re.kr/cgi-bin/decay?Th-234+B-
My understanding is that alpha and beta decays are usually accompanied
by gamma emission because the daughter nucleus is left in an excited
state; it then decays back to its ground state with the emission of
one or more gamma photons. If that's true, I suppose those tables
give the energies and their associated probabilities - except, instead
of probability, they quote something called "intensity." I don't
understand the difference between intensity and probability as used on
that page.
Worse is U-238. It alpha decays to Th-234 with the release of 4.27
MeV. But its page at KAERI also lists "Beta Decay Energy: B- -147.065
+- 1.145 keV" - but if it alpha decays, what the heck is *that* all
about? Moreover, why would the beta decay energy be negative? Is
this an example of inverse beta decay?
Any help at all on this would be immensely appreciated. While the
mere name of a decent text that speaks to nuclear decay in depth would
be wonderful, I'd be glad to read any comments anyone may have to
offer.
--
Dave Typinski
http://home.alltel.net/trapezium
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| User: "David Cross" |
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| Title: Re: Decay chain physics? |
08 Mar 2005 11:36:22 PM |
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"Dave Typinski" <nospam@nospam.net> wrote in message
news:n4us219elos4t8lvpnhos6t1opbglkg4v6@4ax.com...
I'm trying to understand the energy released by (actually, the thermal
power of) nuclear decay chains so I can model geologic radioisotope
decay heat. Anyone care to point me in the right direction with
respect to nuclear decay energy release?
( snip )
Take thorium-234, for example. It beta decays to protactinium-234.
The online table of the nuclides (at KAERI) lists it's decay energy as
273 keV. But, when I click on the "beta" in the page for Th-234, I
can't make heads or tails of the b- and gamma energy distribution
table and energy diagram.
http://atom.kaeri.re.kr/cgi-bin/decay?Th-234+B-
Beta minus or beta plus decay is not monoenergetic so maximum beta energies
are usually given, in case you are not quite clear on that.
My understanding is that alpha and beta decays are usually accompanied
by gamma emission because the daughter nucleus is left in an excited
state; it then decays back to its ground state with the emission of
one or more gamma photons. If that's true, I suppose those tables
give the energies and their associated probabilities - except, instead
of probability, they quote something called "intensity." I don't
understand the difference between intensity and probability as used on
that page.
The intensity is roughly proportional to the height of the gamma-ray peak if
you put a detector on that radioisotope.
I can recommend the intro text by Kenneth Krane, or for a slightly less
daunting overview, try the Ehmann and Vance text "Radiochemistry and Nuclear
Methods of Analysis". I will try to scrounge up some notes I have on basic
energy systematics for the various decays if that would help you. :-)
--
David Cross
dcross1 AT shaw DOT ca
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| User: "Uncle Al" |
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| Title: Re: Decay chain physics? |
09 Mar 2005 11:38:51 AM |
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Dave Typinski wrote:
I'm trying to understand the energy released by (actually, the thermal
power of) nuclear decay chains so I can model geologic radioisotope
decay heat. Anyone care to point me in the right direction with
respect to nuclear decay energy release?
"CRC Handbook" isotope table.
http://t2.lanl.gov/data/astro/molnix96/massd.html
The difference in nuclear GeV mass-equivalent starting nucleus and
product nuclei is the energy yield. That is incremented by electronic
energy differences of a few eV when things settle down. Positron
decay gets you annihalation energy, too, 511 KeV twice.
Yes, yes, I could look up all the relevant geothermodynamic data and
simply grind out an answer; heck, it's probably all been done already
anyway. However, I'm doing this as a recreational diversion and I
want to gain something along the way. Namely, a better understanding
of nuclear decay processes.
Start less end equals difference. Then trim for electronic energies.
I've re-derived the simple equations for a three-step decay chain
that provide the daughter isotopes' quantities with respect to time.
For four step chains and further, I'll appeal to the Bateman
equation... and I haven't yet figured out how to derive that one. I
suppose it wouldn't hurt if I actually bothered to read Bateman's
paper...
Anyway, figuring the power output from any generation with respect to
time is easy *if* one knows the energy released per decay event. Yes,
I know about branching; that's easy to take into account.
Mass difference before and after. Tabulated as above.
Understanding the energy release attendant to a given decay event is
where I'm stuck. Obviously, I need a deeper understanding of nuclear
decay processes. I've searched a good deal online, searched the
_Modern Physics_ text used at UF, and I'm having trouble finding a
good in depth discussion of it all.
Go for the tabulated values.
Take thorium-234, for example. It beta decays to protactinium-234.
The online table of the nuclides (at KAERI) lists it's decay energy as
273 keV. But, when I click on the "beta" in the page for Th-234, I
can't make heads or tails of the b- and gamma energy distribution
table and energy diagram.
http://atom.kaeri.re.kr/cgi-bin/decay?Th-234+B-
Mass difference before and after. Tabulated as above. Trim for
electronic energies. Nuclear events tend to be MeV or thereabouts.
Even tritium decay is tens of KeV. Electronic events are eV, perhaps
hundreds of eV at most for 1s electron capture. Positron (inverse
beta decay) adds twice 511 eV for matter-antimatter annihalation
afterward (e.g., K-40 decay in salt substitute).
My understanding is that alpha and beta decays are usually accompanied
by gamma emission because the daughter nucleus is left in an excited
state; it then decays back to its ground state with the emission of
one or more gamma photons. If that's true, I suppose those tables
give the energies and their associated probabilities - except, instead
of probability, they quote something called "intensity." I don't
understand the difference between intensity and probability as used on
that page.
Worse is U-238. It alpha decays to Th-234 with the release of 4.27
MeV. But its page at KAERI also lists "Beta Decay Energy: B- -147.065
+- 1.145 keV" - but if it alpha decays, what the heck is *that* all
about? Moreover, why would the beta decay energy be negative? Is
this an example of inverse beta decay?
Do it by nuclear mass change. The path is irrelevant. Only start and
end matter.
Any help at all on this would be immensely appreciated. While the
mere name of a decent text that speaks to nuclear decay in depth would
be wonderful, I'd be glad to read any comments anyone may have to
offer.
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/qz.pdf
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| User: "Dave Typinski" |
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| Title: Re: Decay chain physics? |
09 Mar 2005 09:21:06 PM |
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Uncle Al wrote:
Dave Typinski wrote:
I'm trying to understand the energy released by (actually, the thermal
power of) nuclear decay chains so I can model geologic radioisotope
decay heat. Anyone care to point me in the right direction with
respect to nuclear decay energy release?
"CRC Handbook" isotope table.
http://t2.lanl.gov/data/astro/molnix96/massd.html
Nice online ref; thank you.
The difference in nuclear GeV mass-equivalent starting nucleus and
product nuclei is the energy yield.
Got it; thanks again.
That is incremented by electronic
energy differences of a few eV when things settle down.
Yet another side road to explore...
Positron
decay gets you annihalation energy, too, 511 KeV twice.
It does - however, I'd need but half of that. I might not need to
account for any of it.
If a positron is created, the energy to create it and speed it on its
way is accounted for in the mass difference between parent and
daughter nuclei. It's the annihilation of the electron it bumps into
that would have to be added to the total; that's only 511 keV once.
Otoh, if the positron annihilates with a beta already accounted for,
then I can just ignore the whole event.
Otth, it beats me if that last scenario is possible. Pair creation is
another hole I'm in the process of filling.
--
Dave Typinski
http://home.alltel.net/trapezium
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| User: "" |
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| Title: Re: Decay chain physics? |
09 Mar 2005 12:19:22 AM |
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In article <n4us219elos4t8lvpnhos6t1opbglkg4v6@4ax.com>, Dave Typinski <nospam@nospam.net> writes:
I'm trying to understand the energy released by (actually, the thermal
power of) nuclear decay chains so I can model geologic radioisotope
decay heat. Anyone care to point me in the right direction with
respect to nuclear decay energy release?
Yes, yes, I could look up all the relevant geothermodynamic data and
simply grind out an answer; heck, it's probably all been done already
anyway. However, I'm doing this as a recreational diversion and I
want to gain something along the way. Namely, a better understanding
of nuclear decay processes.
I've re-derived the simple equations for a three-step decay chain
that provide the daughter isotopes' quantities with respect to time.
For four step chains and further, I'll appeal to the Bateman
equation... and I haven't yet figured out how to derive that one. I
suppose it wouldn't hurt if I actually bothered to read Bateman's
paper...
Anyway, figuring the power output from any generation with respect to
time is easy *if* one knows the energy released per decay event. Yes,
I know about branching; that's easy to take into account.
Understanding the energy release attendant to a given decay event is
where I'm stuck. Obviously, I need a deeper understanding of nuclear
decay processes. I've searched a good deal online, searched the
_Modern Physics_ text used at UF, and I'm having trouble finding a
good in depth discussion of it all.
Take thorium-234, for example. It beta decays to protactinium-234.
The online table of the nuclides (at KAERI) lists it's decay energy as
273 keV. But, when I click on the "beta" in the page for Th-234, I
can't make heads or tails of the b- and gamma energy distribution
table and energy diagram.
http://atom.kaeri.re.kr/cgi-bin/decay?Th-234+B-
My understanding is that alpha and beta decays are usually accompanied
by gamma emission because the daughter nucleus is left in an excited
state; it then decays back to its ground state with the emission of
one or more gamma photons. If that's true, I suppose those tables
give the energies and their associated probabilities - except, instead
of probability, they quote something called "intensity." I don't
understand the difference between intensity and probability as used on
that page.
Worse is U-238. It alpha decays to Th-234 with the release of 4.27
MeV. But its page at KAERI also lists "Beta Decay Energy: B- -147.065
+- 1.145 keV" - but if it alpha decays, what the heck is *that* all
about? Moreover, why would the beta decay energy be negative? Is
this an example of inverse beta decay?
Any help at all on this would be immensely appreciated. While the
mere name of a decent text that speaks to nuclear decay in depth would
be wonderful, I'd be glad to read any comments anyone may have to
offer.
--
You're getting (for the moment) into too much detail. You can get the
energy of a given decay bycomparing the masses of the parent and
daughter isotopes (delta_e = delta_m*c^2), just make sure you account
for the masses of all the electrons/positrons as well. The fact that,
in the middle you may have an ynholy mess, with different
combinations of particle energies, photons, neutrinos, whatever, is of
no interest for you application, it all thermalizes.
Furthermore, note that the naturally occuring decay chains all follow
the same pattern: you start with a very long lived isotope (lifetime
in the billion of years), then go through a series of intermediate
daughters till you arrive at a stable end point. And, the
intermediate ones are relatively short lived, with lifetimes from
seconds to thousands of years. On a geological time scale this is
immediate. So, you can ignore all the intermediate stages and
directly compare the masses of the initial and final state. As I
recall, both the Uranium and the Thorium chains yield somethin of the
order of 50-60 Mev per nucleon, for the whole chain.
As for the beta decay listed for U238, perhaps they mean the daughter
(Th234) decay here.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
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| User: "Dave Typinski" |
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| Title: Re: Decay chain physics? |
09 Mar 2005 01:39:49 AM |
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wrote:
In article <n4us219elos4t8lvpnhos6t1opbglkg4v6@4ax.com>, Dave Typinski <nospam@nospam.net> writes:
I'm trying to understand the energy released by (actually, the thermal
power of) nuclear decay chains so I can model geologic radioisotope
decay heat. Anyone care to point me in the right direction with
respect to nuclear decay energy release?
...
Understanding the energy release attendant to a given decay event is
where I'm stuck. Obviously, I need a deeper understanding of nuclear
decay processes. I've searched a good deal online, searched the
_Modern Physics_ text used at UF, and I'm having trouble finding a
good in depth discussion of it all.
...
Worse is U-238. It alpha decays to Th-234 with the release of 4.27
MeV. But its page at KAERI also lists "Beta Decay Energy: B- -147.065
+- 1.145 keV" - but if it alpha decays, what the heck is *that* all
about? Moreover, why would the beta decay energy be negative? Is
this an example of inverse beta decay?
You're getting (for the moment) into too much detail.
But... but... the detail is where all the fun is!
You can get the
energy of a given decay bycomparing the masses of the parent and
daughter isotopes (delta_e = delta_m*c^2), ...
D'oh! Yes, of course, that's it. Thank you.
just make sure you account
for the masses of all the electrons/positrons as well. The fact that,
in the middle you may have an ynholy mess, with different
combinations of particle energies, photons, neutrinos, whatever, is of
no interest for you application ...
Yes, in this application.
However, every time I start a project like this, I always find side
roads to take out of curiosity. So, I still want to understand that
region of unholyness, regardless of how impractical that may be.
, it all thermalizes.
Meaning the amount of energy lost to particles and gammas escaping the
lithosphere is trivially small? At least, far enough to the right of
the decimal to not noticeably affect geothermal phenomena?
Furthermore, note that the naturally occuring decay chains all follow
the same pattern: you start with a very long lived isotope (lifetime
in the billion of years), then go through a series of intermediate
daughters till you arrive at a stable end point. And, the
intermediate ones are relatively short lived, with lifetimes from
seconds to thousands of years. On a geological time scale this is
immediate. So, you can ignore all the intermediate stages and
directly compare the masses of the initial and final state. As I
recall, both the Uranium and the Thorium chains yield somethin of the
order of 50-60 Mev per nucleon, for the whole chain.
Aha. Thank you again. I will, of course, work all this out for
myself (not that I doubt you - I just want to prove it to myself) -
but I'm happy to have been pointed in that direction.
As for the beta decay listed for U238, perhaps they mean the daughter
(Th234) decay here.
Here's the page that lists the beta decay energy for U-238, fwiw.
http://atom.kaeri.re.kr/cgi-bin/tonmap12?1206,392
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
--
Dave Typinski
http://home.alltel.net/trapezium
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| User: "Michael Moroney" |
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| Title: Re: Decay chain physics? |
09 Mar 2005 11:56:10 AM |
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Dave Typinski <nospam@nospam.net> writes:
mmeron@cars3.uchicago.edu wrote:
Meaning the amount of energy lost to particles and gammas escaping the
lithosphere is trivially small? At least, far enough to the right of
the decimal to not noticeably affect geothermal phenomena?
Yes. Alphas and betas will get nowhere in solid rock before giving up
all their energy as heat. Gammas don't do much better (several inches on
average)
What you will have to worry about is neutrinos during beta decay. You
can safely assume that 100% of the neutrinos will escape with their energy
and will NOT contribute to heating. While a given beta event will have
a certain fixed amount of energy per decay, it is not evenly divided
among the beta particle and neutrino (and nucleus recoil, which is small).
That is why the beta decay charts will list an average energy per beta
as well as a max. The difference is in the neutrino which escapes.
For a beta that decays to something that gives off a gamma, of course
you can assume 100% of the gamma energy will become heat.
For something that 50% beta decays into the ground state releasing
an average of 0.40 MeV beta energy and 50% into an excited state
averaging 0.20 MeV beta and the resulting gamma is 0.30 MeV you'll
get an average of 0.5*(0.40) + 0.5*(0.20+0.30) or 0.45 MeV.
--
-Mike
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| User: "Dave Typinski" |
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| Title: Re: Decay chain physics? |
10 Mar 2005 01:01:04 AM |
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Michael Moroney wrote:
Dave Typinski <nospam@nospam.net> writes:
mmeron@cars3.uchicago.edu wrote:
Meaning the amount of energy lost to particles and gammas escaping the
lithosphere is trivially small? At least, far enough to the right of
the decimal to not noticeably affect geothermal phenomena?
Yes. Alphas and betas will get nowhere in solid rock before giving up
all their energy as heat. Gammas don't do much better (several inches on
average)
What you will have to worry about is neutrinos during beta decay. You
can safely assume that 100% of the neutrinos will escape with their energy
and will NOT contribute to heating.
Yep. I wonder, though, whether the amount of energy lost to neutrinos
isn't trivial in this context. Currently, planet wide geologic
nuclear decay produces 20 to 60 terrawatts (easy to remember prefix!)
of thermal power depending on the reference in which one looks. How
many watts escape as neutrinos? I'm guessing it's in the kW range -
but, that's the wildest of guesses. So, I'll work it out once I learn
more about it.
While a given beta event will have
a certain fixed amount of energy per decay, it is not evenly divided
among the beta particle and neutrino (and nucleus recoil, which is small).
That is why the beta decay charts will list an average energy per beta
as well as a max. The difference is in the neutrino which escapes.
Is there a statistical relationship between the b- and v energies for
a given nuclear decay? There must be... if it was completely random,
then it *would* be a 50/50 ratio given a large enough number of decay
events.
--
Dave Typinski
http://home.alltel.net/trapezium
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| User: "Michael Moroney" |
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| Title: Re: Decay chain physics? |
10 Mar 2005 01:09:40 PM |
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Dave Typinski <nospam@nospam.net> writes:
What you will have to worry about is neutrinos during beta decay. You
can safely assume that 100% of the neutrinos will escape with their energy
and will NOT contribute to heating.
Yep. I wonder, though, whether the amount of energy lost to neutrinos
isn't trivial in this context. Currently, planet wide geologic
nuclear decay produces 20 to 60 terrawatts (easy to remember prefix!)
of thermal power depending on the reference in which one looks. How
many watts escape as neutrinos? I'm guessing it's in the kW range -
but, that's the wildest of guesses. So, I'll work it out once I learn
more about it.
Umm, no. It's probably closer to half the energy of beta decay is lost
as neutrino energy. Look up on the charts where they list average
beta energy and maximum beta energy. The difference is average neutrino
energy per beta.
I know that for U and Th decay series they are mostly alphas, but there
are a few betas so I'd say several percent will be neutrino energy for
the U and Th decay chains.
However, I know a large contributor to geologic thermal energy is the
decay of K-40, which is a beta decayer. In fact, I think nearly all
EC decay energy is lost as neutrinos, with the small exception being
the small nuclear recoil energy and K X-Rays. A substantial portion of
K-40 decays via EC.
Is there a statistical relationship between the b- and v energies for
a given nuclear decay? There must be... if it was completely random,
then it *would* be a 50/50 ratio given a large enough number of decay
events.
I've seen graphs that show the beta energy vs. frequency. They seem to
all start off at 0% for 0 beta energy (100% into neutrino energy) rise
to a peak nearish (but not at) 50% beta/50% neutrino energy, then
decrease to zero at 100% beta/0% neutrino energy. As I mentioned, the
charts give an average and maximum beta energy. The difference is the
average (lost) neutrino energy. For example for one of the Th-234
decays, the chart shows the max beta energy is 198.5 keV, while the
average is only 52.7 keV. Meaning that 145.8 keV is average neutrino
energy.
--
-Mike
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| User: "Dave Typinski" |
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| Title: Re: Decay chain physics? |
10 Mar 2005 06:10:58 PM |
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Michael Moroney wrote:
Dave Typinski <nospam@nospam.net> writes:
What you will have to worry about is neutrinos during beta decay. You
can safely assume that 100% of the neutrinos will escape with their energy
and will NOT contribute to heating.
Yep. I wonder, though, whether the amount of energy lost to neutrinos
isn't trivial in this context. Currently, planet wide geologic
nuclear decay produces 20 to 60 terrawatts (easy to remember prefix!)
of thermal power depending on the reference in which one looks. How
many watts escape as neutrinos? I'm guessing it's in the kW range -
but, that's the wildest of guesses. So, I'll work it out once I learn
more about it.
Umm, no. It's probably closer to half the energy of beta decay is lost
as neutrino energy. Look up on the charts where they list average
beta energy and maximum beta energy. The difference is average neutrino
energy per beta.
I know that for U and Th decay series they are mostly alphas, but there
are a few betas so I'd say several percent will be neutrino energy for
the U and Th decay chains.
However, I know a large contributor to geologic thermal energy is the
decay of K-40, which is a beta decayer. In fact, I think nearly all
EC decay energy is lost as neutrinos, with the small exception being
the small nuclear recoil energy and K X-Rays. A substantial portion of
K-40 decays via EC.
Gasp! That means bananas are really super secret government mind
control neutrino emitters! The horror, the horror... I just *knew*
there had to be an ulterior motive behind U. S. involvement in Central
America. Chiquita is a front and Iran-Contra was nothing but a
subterfuge to distract us from the hideous truth: banana based mind
manipulation.
That would certainly explain some of the weirder pseudoscientific
stuff the Soviets seemed to take seriously: they were simply worried
about a banana gap.
Seriously, though, I've always wondered whether the Soviet dalliance
with ESP and telekinesis was approved by the politburo or was simply a
coupla loose canon True Believers. I'm pretty sure American (CIA)
projects along the same lines were a bit of both.
Is there a statistical relationship between the b- and v energies for
a given nuclear decay? There must be... if it was completely random,
then it *would* be a 50/50 ratio given a large enough number of decay
events.
I've seen graphs that show the beta energy vs. frequency. They seem to
all start off at 0% for 0 beta energy (100% into neutrino energy) rise
to a peak nearish (but not at) 50% beta/50% neutrino energy, then
decrease to zero at 100% beta/0% neutrino energy. As I mentioned, the
charts give an average and maximum beta energy. The difference is the
average (lost) neutrino energy. For example for one of the Th-234
decays, the chart shows the max beta energy is 198.5 keV, while the
average is only 52.7 keV. Meaning that 145.8 keV is average neutrino
energy.
That helps; thank you.
--
Dave Typinski
http://home.alltel.net/trapezium
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| User: "Puppet_Sock" |
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| Title: Re: Decay chain physics? |
10 Mar 2005 12:01:07 PM |
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Dave Typinski wrote:
[regarding beta decay and the energy carried off by neutrino]
Is there a statistical relationship between the b- and v energies for
a given nuclear decay? There must be... if it was completely random,
then it *would* be a 50/50 ratio given a large enough number of decay
events.
It is random, but according to a pattern. Put fairly simplisticly,
it goes like so. You start with a nuclide more or less at rest.
You end up with a different nuclide, with an electric charge
different by 1, a beta particle, and a neutrino, all flying away
from the centre. The total momentum has to be zero, as it started
zero. But which particle goes which direction is random. And that's
where the spectrum of the neutrino energy comes from.
You can look up fairly good averages. For example, you could find
it in one of the early chapters of Lamarsh _Intro to Nuclear
Engineering_. For geological situations, the average won't be
affected by anything going on in ordinary situations for most
important isotopes.
You should look at a CRC handbook isotopes table. You can find
the half lives and decay energies for pretty much all important
isotopes. If you then know how much of each isotope there is in
your model, it's quite direct to calculate such things as the
energy output.
Socks
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| User: "Dave Typinski" |
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| Title: Re: Decay chain physics? |
10 Mar 2005 05:43:24 PM |
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Puppet_Sock wrote:
Dave Typinski wrote:
[regarding beta decay and the energy carried off by neutrino]
Is there a statistical relationship between the b- and v energies for
a given nuclear decay? There must be... if it was completely random,
then it *would* be a 50/50 ratio given a large enough number of decay
events.
It is random, but according to a pattern. Put fairly simplisticly,
it goes like so. You start with a nuclide more or less at rest.
You end up with a different nuclide, with an electric charge
different by 1, a beta particle, and a neutrino, all flying away
from the centre. The total momentum has to be zero, as it started
zero. But which particle goes which direction is random. And that's
where the spectrum of the neutrino energy comes from.
Is it the energy or the direction that's random? Or both? Is there
even a way to tell? That is, is there a causal relationship here?
If the neutrino grabs, say, 30% of the beta decay energy, then the
beta gets almost 70%, with the remainder going into the recoil of the
nucleus. In that case, there's at most two directions in which the
neutrino could move to conserve momentum.
Otoh, if the neutrino goes off in a certain direction relative to the
nucleus and the beta, then there's only one specific energy it can
have for given momenta of the beta and the nucleus.
Maybe a better question is, since it all boils down to the same
result, does it even matter? I suspect these are simply two ways of
looking at the same phenomena.
--
Dave Typinski
http://home.alltel.net/trapezium
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| User: "David Cross" |
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| Title: Re: Decay chain physics? |
10 Mar 2005 06:03:16 PM |
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On Thu, 10 Mar 2005 18:43:24 -0500, Dave Typinski <nospam@nospam.net> wrote:
Is it the energy or the direction that's random? Or both? Is there
even a way to tell? That is, is there a causal relationship here?
The direction is (should be) isotropic. In fact, under carefully controlled
conditions you can produce an asymmetry in distribution of the beta particles.
This is due to the nonconservation of parity in the beta decay, but not
important for your purposes.
What is important is that the total energy of the decay is split in
statistical fashion between the neutrino and the beta particle, and your major
concern is the E-beta-max, since that's the total energy of decay where the
neutrino gets no kinetic energy at all and the electron or positron gets all
of it, and thus the total energy of decay one way or another goes into the
Earth.
A rule of thumb is that the E-beta-average is around one-third to one-half
E-beta-max, and so if you are concerned that neutrino emissions reduce the
energy deposited into the Earth's crust or mantle, then if you want to
approximate the energies given off in the beta decays of the uranium and
thorium decay chains, then use the rule of thumb above.
---
David Cross
dcross1 AT shaw DOT ca
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| User: "" |
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| Title: Re: Decay chain physics? |
10 Mar 2005 01:43:21 AM |
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In article <15sv21dj4a6cac5m82pn47sth773r2fvaj@4ax.com>, Dave Typinski <nospam@nospam.net> writes:
Michael Moroney wrote:
Dave Typinski <nospam@nospam.net> writes:
mmeron@cars3.uchicago.edu wrote:
Meaning the amount of energy lost to particles and gammas escaping the
lithosphere is trivially small? At least, far enough to the right of
the decimal to not noticeably affect geothermal phenomena?
Yes. Alphas and betas will get nowhere in solid rock before giving up
all their energy as heat. Gammas don't do much better (several inches on
average)
What you will have to worry about is neutrinos during beta decay. You
can safely assume that 100% of the neutrinos will escape with their energy
and will NOT contribute to heating.
Yep. I wonder, though, whether the amount of energy lost to neutrinos
isn't trivial in this context.
Not quite.
Currently, planet wide geologic
nuclear decay produces 20 to 60 terrawatts (easy to remember prefix!)
of thermal power depending on the reference in which one looks. How
many watts escape as neutrinos? I'm guessing it's in the kW range -
but, that's the wildest of guesses. So, I'll work it out once I learn
more about it.
The energy taken by neutrinos will be few percent of the total.
Roughly (very roughly, it has been years since I looked at this
stuff), close to 90% of the energy of the decay chain comes in the
form of alphas and the associated gammas (emitted when the alpha decay
doesn't go to ground state). The rest is betas and a significant
fraction of this goes to the neutrinos. So, few percent of your 20-60
terawats.
While a given beta event will have
a certain fixed amount of energy per decay, it is not evenly divided
among the beta particle and neutrino (and nucleus recoil, which is small).
That is why the beta decay charts will list an average energy per beta
as well as a max. The difference is in the neutrino which escapes.
Is there a statistical relationship between the b- and v energies for
a given nuclear decay? There must be... if it was completely random,
then it *would* be a 50/50 ratio given a large enough number of decay
events.
There are such relationships, essentially it is a matter of
kinematics. I'm sure you'll be able to find them.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
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| User: "Dave Typinski" |
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| Title: Re: Decay chain physics? |
10 Mar 2005 05:13:41 PM |
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wrote:
In article <15sv21dj4a6cac5m82pn47sth773r2fvaj@4ax.com>, Dave Typinski <nospam@nospam.net> writes:
Michael Moroney wrote:
Dave Typinski <nospam@nospam.net> writes:
wrote:
Meaning the amount of energy lost to particles and gammas escaping the
lithosphere is trivially small? At least, far enough to the right of
the decimal to not noticeably affect geothermal phenomena?
Yes. Alphas and betas will get nowhere in solid rock before giving up
all their energy as heat. Gammas don't do much better (several inches on
average)
What you will have to worry about is neutrinos during beta decay. You
can safely assume that 100% of the neutrinos will escape with their energy
and will NOT contribute to heating.
Yep. I wonder, though, whether the amount of energy lost to neutrinos
isn't trivial in this context.
Not quite.
Currently, planet wide geologic
nuclear decay produces 20 to 60 terrawatts (easy to remember prefix!)
of thermal power depending on the reference in which one looks. How
many watts escape as neutrinos? I'm guessing it's in the kW range -
but, that's the wildest of guesses. So, I'll work it out once I learn
more about it.
The energy taken by neutrinos will be few percent of the total.
Roughly (very roughly, it has been years since I looked at this
stuff), close to 90% of the energy of the decay chain comes in the
form of alphas and the associated gammas (emitted when the alpha decay
doesn't go to ground state). The rest is betas and a significant
fraction of this goes to the neutrinos. So, few percent of your 20-60
terawats.
Wow, that much? That's definitely worth consideration here.
Given a neutrino mass of 1 eV or less (or has that been narrowed
down?), that's one whopping lot of neutrinos. Or a somewhat lesser
amount of really, really, really fast neutrinos. It's a good thing
they don't interact with much of anything... what's the absorption
length in lead of solar neutrinos, a parsec or so?
While a given beta event will have
a certain fixed amount of energy per decay, it is not evenly divided
among the beta particle and neutrino (and nucleus recoil, which is small).
That is why the beta decay charts will list an average energy per beta
as well as a max. The difference is in the neutrino which escapes.
Is there a statistical relationship between the b- and v energies for
a given nuclear decay? There must be... if it was completely random,
then it *would* be a 50/50 ratio given a large enough number of decay
events.
There are such relationships, essentially it is a matter of
kinematics. I'm sure you'll be able to find them.
You bet I will. Thanks.
--
Dave Typinski
http://home.alltel.net/trapezium
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| User: "" |
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| Title: Re: Decay chain physics? |
10 Mar 2005 07:37:27 PM |
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In article <mfk131tked96s8js1hiqi7av2v4k8lj50q@4ax.com>, Dave Typinski <nospam@nospam.net> writes:
mmeron@cars3.uchicago.edu wrote:
In article <15sv21dj4a6cac5m82pn47sth773r2fvaj@4ax.com>, Dave Typinski <nospam@nospam.net> writes:
Michael Moroney wrote:
Dave Typinski <nospam@nospam.net> writes:
mmeron@cars3.uchicago.edu wrote:
Meaning the amount of energy lost to particles and gammas escaping the
lithosphere is trivially small? At least, far enough to the right of
the decimal to not noticeably affect geothermal phenomena?
Yes. Alphas and betas will get nowhere in solid rock before giving up
all their energy as heat. Gammas don't do much better (several inches on
average)
What you will have to worry about is neutrinos during beta decay. You
can safely assume that 100% of the neutrinos will escape with their energy
and will NOT contribute to heating.
Yep. I wonder, though, whether the amount of energy lost to neutrinos
isn't trivial in this context.
Not quite.
Currently, planet wide geologic
nuclear decay produces 20 to 60 terrawatts (easy to remember prefix!)
of thermal power depending on the reference in which one looks. How
many watts escape as neutrinos? I'm guessing it's in the kW range -
but, that's the wildest of guesses. So, I'll work it out once I learn
more about it.
The energy taken by neutrinos will be few percent of the total.
Roughly (very roughly, it has been years since I looked at this
stuff), close to 90% of the energy of the decay chain comes in the
form of alphas and the associated gammas (emitted when the alpha decay
doesn't go to ground state). The rest is betas and a significant
fraction of this goes to the neutrinos. So, few percent of your 20-60
terawats.
Wow, that much? That's definitely worth consideration here.
Given a neutrino mass of 1 eV or less (or has that been narrowed
down?), that's one whopping lot of neutrinos.
Their mass is not relevant in this case, just their energy.
Or a somewhat lesser amount of really, really, really fast neutrinos.
Well, yeas, one neutrino per beta decay.
It's a good thing
they don't interact with much of anything... what's the absorption
length in lead of solar neutrinos, a parsec or so?
Depending on energy but something of this order, yes.
While a given beta event will have
a certain fixed amount of energy per decay, it is not evenly divided
among the beta particle and neutrino (and nucleus recoil, which is small).
That is why the beta decay charts will list an average energy per beta
as well as a max. The difference is in the neutrino which escapes.
Is there a statistical relationship between the b- and v energies for
a given nuclear decay? There must be... if it was completely random,
then it *would* be a 50/50 ratio given a large enough number of decay
events.
There are such relationships, essentially it is a matter of
kinematics. I'm sure you'll be able to find them.
You bet I will. Thanks.
Good luck
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
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| User: "" |
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| Title: Re: Decay chain physics? |
09 Mar 2005 03:04:51 AM |
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In article <m18t21trhabape1ndep93bf0ohjhi6p7lc@4ax.com>, Dave Typinski <nospam@nospam.net> writes:
mmeron@cars3.uchicago.edu wrote:
In article <n4us219elos4t8lvpnhos6t1opbglkg4v6@4ax.com>, Dave Typinski <nospam@nospam.net> writes:
I'm trying to understand the energy released by (actually, the thermal
power of) nuclear decay chains so I can model geologic radioisotope
decay heat. Anyone care to point me in the right direction with
respect to nuclear decay energy release?
...
Understanding the energy release attendant to a given decay event is
where I'm stuck. Obviously, I need a deeper understanding of nuclear
decay processes. I've searched a good deal online, searched the
_Modern Physics_ text used at UF, and I'm having trouble finding a
good in depth discussion of it all.
...
Worse is U-238. It alpha decays to Th-234 with the release of 4.27
MeV. But its page at KAERI also lists "Beta Decay Energy: B- -147.065
+- 1.145 keV" - but if it alpha decays, what the heck is *that* all
about? Moreover, why would the beta decay energy be negative? Is
this an example of inverse beta decay?
You're getting (for the moment) into too much detail.
But... but... the detail is where all the fun is!
Oh, sure, no argument about it.
You can get the
energy of a given decay bycomparing the masses of the parent and
daughter isotopes (delta_e = delta_m*c^2), ...
D'oh! Yes, of course, that's it. Thank you.
just make sure you account
for the masses of all the electrons/positrons as well. The fact that,
in the middle you may have an ynholy mess, with different
combinations of particle energies, photons, neutrinos, whatever, is of
no interest for you application ...
Yes, in this application.
However, every time I start a project like this, I always find side
roads to take out of curiosity. So, I still want to understand that
region of unholyness, regardless of how impractical that may be.
Well, I always found following all possible tangents to be a most
satisfying form of procrastination, so I can fully identify with how
you feel:-)
, it all thermalizes.
Meaning the amount of energy lost to particles and gammas escaping the
lithosphere is trivially small? At least, far enough to the right of
the decimal to not noticeably affect geothermal phenomena?
Oh, certainly. Even for the most energetic nuclear gammas the
absorption length (the thickness within which the intensity declines
by a factor of e) in rock/soil is of the order of tens of centimeters.
For alphas or betas you're talking about microns. So, yes, the amount
escaping is trivially small. Consider that the shielding around a
nuclear reactor (couple meters of concrete or so) suffices to reduce
the radiation levels by many orders of magnitude.
Furthermore, note that the naturally occuring decay chains all follow
the same pattern: you start with a very long lived isotope (lifetime
in the billion of years), then go through a series of intermediate
daughters till you arrive at a stable end point. And, the
intermediate ones are relatively short lived, with lifetimes from
seconds to thousands of years. On a geological time scale this is
immediate. So, you can ignore all the intermediate stages and
directly compare the masses of the initial and final state. As I
recall, both the Uranium and the Thorium chains yield somethin of the
order of 50-60 Mev per nucleon, for the whole chain.
Aha. Thank you again. I will, of course, work all this out for
myself
Of course, you should.
(not that I doubt you - I just want to prove it to myself) -
but I'm happy to have been pointed in that direction.
As for the beta decay listed for U238, perhaps they mean the daughter
(Th234) decay here.
Here's the page that lists the beta decay energy for U-238, fwiw.
http://atom.kaeri.re.kr/cgi-bin/tonmap12?1206,392
Well, I've no idea what they mean by this. U-238 itself has no beta
decays.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
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| User: "Y.Porat" |
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| Title: Re: Decay chain physics? |
09 Mar 2005 01:19:11 AM |
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you might fing something interesting in the thread that was going here
not long ago:
'isotope decay chains'
and that is only the 'tip of the iceberg'
all the ebst
Y.Porat
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